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Minimum Array End

Medium

Asked by:

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Topics:

Bit Manipulation

You are given two integers n and x. You have to construct an array of positive integers nums of size n where for every 0 <= i < n - 1, nums[i + 1] is greater than nums[i], and the result of the bitwise AND operation between all elements of nums is x.

Return the minimum possible value of nums[n - 1].

Example 1:

Input: n = 3, x = 4

Output: 6

Explanation:

nums can be [4,5,6] and its last element is 6.

Example 2:

Input: n = 2, x = 7

Output: 15

Explanation:

nums can be [7,15] and its last element is 15.

Constraints:

1 <= n, x <= 10⁸

Try coding it in LeetCode

Minimum Array End

Medium

Asked by:

18 views

Topics:

Bit Manipulation

Return the minimum possible value of nums[n - 1].

Example 1:

Input: n = 3, x = 4

Output: 6

Explanation:

nums can be [4,5,6] and its last element is 6.

Example 2:

Input: n = 2, x = 7

Output: 15

Explanation:

nums can be [7,15] and its last element is 15.

Constraints:

1 <= n, x <= 10⁸

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the expected data type of the elements in the array, and what is the range of possible values for each element?
Can the input array be empty, or contain null or undefined values? If so, what should be the expected output?
Are there any specific conditions under which a 'minimum end' would not exist, and if so, what should I return in that case?
Is the array sorted in any particular order (ascending, descending, or unsorted), and does the definition of 'minimum end' depend on whether it's sorted?
Could you provide a more concrete example of what constitutes a 'minimum end' in various scenarios, especially with edge cases or duplicate values?

Brute Force Solution

Approach

The brute force method involves exploring every single possible arrangement to find the minimum. It's like trying out all combinations until you stumble upon the right one. This involves checking each possibility to ensure it meets the required conditions and then choosing the best one.

Here's how the algorithm would work step-by-step:

Consider all possible start points in the array.
For each start point, check every possible length of the sub-array starting from that point.
Evaluate each possible sub-array based on your target criteria.
Keep track of the best outcome (minimum value at the end) found so far.
Continue until you've considered all possible sub-arrays.
Return the best outcome found from all the evaluations.

Code Implementation

def find_minimum_array_end_brute_force(input_array):
    minimum_end_value = float('inf')

    # Iterate through all possible start indices
    for start_index in range(len(input_array)):
        # Iterate through all possible lengths of sub-arrays
        for sub_array_length in range(1, len(input_array) - start_index + 1):

            current_sub_array = input_array[start_index:start_index + sub_array_length]

            # Update minimum if current end value is smaller
            if current_sub_array[-1] < minimum_end_value:

                minimum_end_value = current_sub_array[-1]

    # Handle the edge case of an empty input array
    if not input_array:
        return None

    return minimum_end_value

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through all possible start indices in the array, which contributes a factor of 'n'. For each start index, it considers all possible lengths of the sub-array, up to the end of the array, contributing another factor related to 'n'. Therefore, the algorithm performs roughly n * n operations in the worst-case scenario, where 'n' is the size of the input array. This means the time complexity can be approximated as n * n, which simplifies to O(n²).

Space Complexity

O(1) – The provided brute force approach iterates through all possible start points and lengths of sub-arrays, evaluating each based on the target criteria and keeping track of the best outcome found so far. This involves storing only a few variables: loop counters, temporary variables to hold the current sub-array's value, and the best outcome (minimum end value) found so far. The number of variables used does not depend on the input array size N, meaning that the space complexity is constant. Therefore, the space complexity is O(1).

Optimal Solution

Approach

The goal is to find the smallest possible number that can be left at the end of a series of operations. Instead of trying all possible combinations of operations, we focus on strategically reducing the numbers step-by-step. This ensures we arrive at the absolute minimum possible value.

Here's how the algorithm would work step-by-step:

Start by thinking about what the very last operation must do to achieve the smallest end result.
Realize that the last operation will either add or multiply some numbers. Focus on arranging the numbers to make this last step result in the lowest value.
Consider whether addition or multiplication results in a smaller number at each stage.
Strategically work backwards, optimizing each step to make sure the final operation results in the smallest number. This prevents needing to check every order of operations.
The numbers should be arranged in a way that allows the lowest numbers to be handled last.

Code Implementation

def find_minimum_array_end(input_numbers): 
    input_numbers.sort(reverse=True)
    current_result = input_numbers[0]

    # Iterate through the remaining numbers
    for i in range(1, len(input_numbers)): 
        # Prioritize multiplication when the current number is greater than 1.
        if input_numbers[i] > 1:
            current_result *= input_numbers[i]

        # Otherwise, add to try and reach a minimal value
        else:
            current_result += input_numbers[i]

    return current_result

Big(O) Analysis

Time Complexity

O(n log n) – The suggested approach involves strategically arranging and optimizing the numbers step-by-step, aiming to minimize the final result without exhaustively checking all permutations. Implicitly, this includes sorting the numbers to facilitate the optimal arrangement for minimizing the end result, allowing lower numbers to be handled last. Sorting the array of n elements takes O(n log n) time. Subsequent operations to add or multiply will occur in each step but will happen in at most n operations, making it a lower order term. Thus, the dominant factor is O(n log n).

Space Complexity

O(1) – The plain English explanation focuses on strategically arranging and operating on numbers in place, with the core idea being to optimize steps without exploring all permutations. This suggests that the operations primarily involve modifying the original array directly, rather than creating auxiliary data structures that scale with the input size. The explanation does not mention the use of any temporary arrays, hash maps, or recursion. Therefore, the space complexity is likely constant, as only a few variables may be needed for tracking the current state or intermediate results, independent of the input size N.

Edge Cases

Case	How to Handle
Null or undefined input array	Throw an IllegalArgumentException or return null/empty list after checking for null input to prevent NullPointerException.
Array with only one element	Return an empty list since at least two elements are needed to perform the minimum array end operations.
All elements in the array are identical	The algorithm should still perform correctly, potentially requiring more iterations if swaps are necessary for the minimum end condition.
Array with negative numbers	The algorithm should function correctly with negative numbers as the core logic is based on comparing and moving elements.
Large array size exceeding memory limits	Consider a more space-efficient in-place algorithm or using disk-based processing if memory is a significant constraint.
Integer overflow during comparison or calculations	Use long data type to prevent overflow when checking/using integer data, or implement checks for potential overflow during calculations.
Array is already in the minimum array end state	The algorithm should identify this and terminate quickly to avoid unnecessary computations, returning the original array.
Array with extreme boundary values (e.g., Integer.MAX_VALUE, Integer.MIN_VALUE)	The comparison logic needs to correctly handle extreme values and avoid overflow or incorrect ordering behavior.

Null or undefined input array

How to Handle:

Throw an IllegalArgumentException or return null/empty list after checking for null input to prevent NullPointerException.

Array with only one element

How to Handle:

Return an empty list since at least two elements are needed to perform the minimum array end operations.

All elements in the array are identical

How to Handle:

The algorithm should still perform correctly, potentially requiring more iterations if swaps are necessary for the minimum end condition.

Array with negative numbers

How to Handle:

The algorithm should function correctly with negative numbers as the core logic is based on comparing and moving elements.

Large array size exceeding memory limits

How to Handle:

Consider a more space-efficient in-place algorithm or using disk-based processing if memory is a significant constraint.

Integer overflow during comparison or calculations

How to Handle:

Use long data type to prevent overflow when checking/using integer data, or implement checks for potential overflow during calculations.

Array is already in the minimum array end state

How to Handle:

The algorithm should identify this and terminate quickly to avoid unnecessary computations, returning the original array.

Array with extreme boundary values (e.g., Integer.MAX_VALUE, Integer.MIN_VALUE)

How to Handle:

The comparison logic needs to correctly handle extreme values and avoid overflow or incorrect ordering behavior.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the expected data type of the elements in the array, and what is the range of possible values for each element?
Can the input array be empty, or contain null or undefined values? If so, what should be the expected output?
Are there any specific conditions under which a 'minimum end' would not exist, and if so, what should I return in that case?
Is the array sorted in any particular order (ascending, descending, or unsorted), and does the definition of 'minimum end' depend on whether it's sorted?
Could you provide a more concrete example of what constitutes a 'minimum end' in various scenarios, especially with edge cases or duplicate values?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Consider all possible start points in the array.
For each start point, check every possible length of the sub-array starting from that point.
Evaluate each possible sub-array based on your target criteria.
Keep track of the best outcome (minimum value at the end) found so far.
Continue until you've considered all possible sub-arrays.
Return the best outcome found from all the evaluations.

Code Implementation

def find_minimum_array_end_brute_force(input_array):
    minimum_end_value = float('inf')

    # Iterate through all possible start indices
    for start_index in range(len(input_array)):
        # Iterate through all possible lengths of sub-arrays
        for sub_array_length in range(1, len(input_array) - start_index + 1):

            current_sub_array = input_array[start_index:start_index + sub_array_length]

            # Update minimum if current end value is smaller
            if current_sub_array[-1] < minimum_end_value:

                minimum_end_value = current_sub_array[-1]

    # Handle the edge case of an empty input array
    if not input_array:
        return None

    return minimum_end_value

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Start by thinking about what the very last operation must do to achieve the smallest end result.
Realize that the last operation will either add or multiply some numbers. Focus on arranging the numbers to make this last step result in the lowest value.
Consider whether addition or multiplication results in a smaller number at each stage.
Strategically work backwards, optimizing each step to make sure the final operation results in the smallest number. This prevents needing to check every order of operations.
The numbers should be arranged in a way that allows the lowest numbers to be handled last.

Code Implementation

def find_minimum_array_end(input_numbers): 
    input_numbers.sort(reverse=True)
    current_result = input_numbers[0]

    # Iterate through the remaining numbers
    for i in range(1, len(input_numbers)): 
        # Prioritize multiplication when the current number is greater than 1.
        if input_numbers[i] > 1:
            current_result *= input_numbers[i]

        # Otherwise, add to try and reach a minimal value
        else:
            current_result += input_numbers[i]

    return current_result

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or undefined input array	Throw an IllegalArgumentException or return null/empty list after checking for null input to prevent NullPointerException.
Array with only one element	Return an empty list since at least two elements are needed to perform the minimum array end operations.
All elements in the array are identical	The algorithm should still perform correctly, potentially requiring more iterations if swaps are necessary for the minimum end condition.
Array with negative numbers	The algorithm should function correctly with negative numbers as the core logic is based on comparing and moving elements.
Large array size exceeding memory limits	Consider a more space-efficient in-place algorithm or using disk-based processing if memory is a significant constraint.
Integer overflow during comparison or calculations	Use long data type to prevent overflow when checking/using integer data, or implement checks for potential overflow during calculations.
Array is already in the minimum array end state	The algorithm should identify this and terminate quickly to avoid unnecessary computations, returning the original array.
Array with extreme boundary values (e.g., Integer.MAX_VALUE, Integer.MIN_VALUE)	The comparison logic needs to correctly handle extreme values and avoid overflow or incorrect ordering behavior.

Null or undefined input array

How to Handle:

Throw an IllegalArgumentException or return null/empty list after checking for null input to prevent NullPointerException.

Array with only one element

How to Handle:

Return an empty list since at least two elements are needed to perform the minimum array end operations.

All elements in the array are identical

How to Handle:

The algorithm should still perform correctly, potentially requiring more iterations if swaps are necessary for the minimum end condition.

Array with negative numbers

How to Handle:

The algorithm should function correctly with negative numbers as the core logic is based on comparing and moving elements.

Large array size exceeding memory limits

How to Handle:

Consider a more space-efficient in-place algorithm or using disk-based processing if memory is a significant constraint.

Integer overflow during comparison or calculations

How to Handle:

Use long data type to prevent overflow when checking/using integer data, or implement checks for potential overflow during calculations.

Array is already in the minimum array end state

How to Handle:

The algorithm should identify this and terminate quickly to avoid unnecessary computations, returning the original array.

Array with extreme boundary values (e.g., Integer.MAX_VALUE, Integer.MIN_VALUE)

How to Handle:

The comparison logic needs to correctly handle extreme values and avoid overflow or incorrect ordering behavior.